Question

Suppose $\left|\begin{array}{cc}{f}^{\prime }\left(x\right)& f\left(x\right)\\ f^{\prime \prime }\left(x\right)& f^{\prime }\left(x\right)\end{array}\right|=0$ where $f\left(x\right)$ is continuously differentiable function with $f^{\prime }\left(x\right)\ne 0$ and satisfies $f\left(0\right)=1$ and $f^{\prime }\left(0\right)=2$, then $f\left(x\right)$ is

• A. $x^2+2x+1$
• B. $2e^x-1$
• C. $e^{2x}$
• D. $4e^{\frac{x}{2}}-3$

Answer 1

The correct option is C $e^{2x}$

Solving the given determinant, we get, $f^{\prime }(x{)}^2=f^{\prime \prime }(x\left)f\right(x)$

$\Rightarrow \frac{f^{\prime \prime }\left(x\right)}{f^{\prime }\left(x\right)}=\frac{f^{\prime }\left(x\right)}{f\left(x\right)}$

Integrating on both sides,

$\Rightarrow f^{\prime }\left(x\right)=k\times f\left(x\right)$

Since $f\left(0\right)=1$ and $f^{\prime }\left(0\right)=2$ $\Rightarrow k=2$

$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=2$

Now integrating on both sides,

$\Rightarrow \ln \left(f\right(x\left)\right)=2k^{\prime }\times x$

$\Rightarrow f\left(x\right)=e^{2k^{\prime }x}$

$f^{\prime }\left(0\right)=2$ $\Rightarrow$ $k^{\prime }=1$

So, option C is correct.